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%I #11 Apr 05 2018 09:06:06
%S 4,35,446,2827,13686,59859,250198,1023347,4140830,16663627,66867438,
%T 267922683,1072654438,4292666147,17175010598,68709242467,274856398542,
%U 1099466524251,4397952122110,17591988892331,70368333113174
%N Number of arrangements of n+2 numbers in 0..3 with each number being the sum mod 4 of two others.
%C Column 3 of A183884.
%H R. H. Hardin, <a href="/A183878/b183878.txt">Table of n, a(n) for n = 1..199</a>
%F Empirical (for n>=3): 4^(n+2) - (2*n+7)*2^(n+2) - 2*n^3 - 9*n^2 - 10*n + 3. - _Vaclav Kotesovec_, Nov 27 2012
%F Conjectures from _Colin Barker_, Apr 05 2018: (Start)
%F G.f.: x*(4 - 13*x + 258*x^2 - 1087*x^3 + 1318*x^4 + 444*x^5 - 2040*x^6 + 1536*x^7 - 384*x^8) / ((1 - x)^4*(1 - 2*x)^2*(1 - 4*x)).
%F a(n) = 12*a(n-1) - 58*a(n-2) + 148*a(n-3) - 217*a(n-4) + 184*a(n-5) - 84*a(n-6) + 16*a(n-7) for n>9.
%F (End)
%e Some solutions for n=2:
%e ..1....2....0....0....2....2....3....1....2....2....2....0....3....0....2....2
%e ..3....0....2....2....1....3....1....3....2....1....0....2....2....0....2....2
%e ..3....0....2....2....1....1....1....1....2....3....2....0....1....2....0....0
%e ..2....2....2....0....3....3....2....2....0....1....2....2....3....2....2....0
%Y Cf. A183884.
%K nonn
%O 1,1
%A _R. H. Hardin_, Jan 07 2011
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